First screening kernel from a finite symmetric-polynomial calculation

Put $q=(p-1)\mathrm{/}(2p)$ and suppress charge labels on polynomials. The oscillator formula for $L(-k)$ on charge $\beta$ is $L(-k)={\sum }_{j\ge 1}j{x}_{k+j}{\mathrm{\partial }}_{x_j}+\frac{1}{4p}{\sum }_{j=1}^{k-1}{x}_j{x}_{k-j}+(\beta +q(k-1))x_k\mathrm{.}$ On the charge $1\mathrm{/}2$ Fock space, its positive mode is $L(k)={\sum }_{j\ge 1}(k+j)x_j{\mathrm{\partial }}_{x_{k+j}}+p{\sum }_{j=1}^{k-1}j(k-j){\mathrm{\partial }}_{x_j}{\mathrm{\partial }}_{x_{k-j}}+k(1-(p-1)k){\mathrm{\partial }}_{x_k}\mathrm{.}$ Define a nondegenerate bilinear pairing between each equal-degree pair of these two polynomial spaces by making distinct monomials orthogonal and setting $⟨\prod x_j^{m_j},\prod x_j^{m_j}⟩=\prod m_j!(-2pj{)}^{m_j}$. Thus the adjoint of multiplication by $x_j$ is $-2pj{\mathrm{\partial }}_{x_j}$. Applying this rule termwise to the negative-mode formula gives exactly the positive-mode formula: its linear coefficient is $-2pk(\beta +q(k-1))=k(1-(p-1)k)$. Consequently $⟨L(-k)f,g⟩=⟨f,L(k)g⟩$ whenever the degrees match. These are finite-dimensional algebraic pairings and require no analytic completion.

On the charge-$1\mathrm{/}2$ polynomial space form the degree-preserving finite-on-each-degree operator $H={\sum }_{k\ge 1}{x}_{k}L(k)$. Put $\mathcal{E}={\sum }_{k\ge 1}k{x}_k{\mathrm{\partial }}_{x_k}$ and define $J_p={\sum }_{a,b\ge 1}(a+b)x_a{x}_b{\mathrm{\partial }}_{x_{a+b}}+p{\sum }_{a,b\ge 1}ab{x}_{a+b}{\mathrm{\partial }}_{x_a}{\mathrm{\partial }}_{x_b}+(p-1){\sum }_{k\ge 1}k(k-1)x_k{\mathrm{\partial }}_{x_k}\mathrm{.}$ Let $\sigma$ replace every $x_k$ by $-x_k$. The two cubic terms change sign under conjugation by $\sigma$, whereas $x_k{\mathrm{\partial }}_{x_k}$ does not. The positive-mode formula of step 1.1 therefore gives the exact identity $\sigma H\sigma =-J_p+(2-p)\mathcal{E}$.

Fix $n\ge 1$ and choose an integer $m\ge n$. Substitution $x_k={\sum }_{i=1}^m{y}_i^k$ identifies $P_n$ with the degree-$n$ symmetric polynomials in $m$ variables. Indeed the first $m$ power sums generate the elementary symmetric polynomials by the recursion $r{e}_r={\sum }_{j=1}^r(-1{)}^{j-1}{e}_{r-j}{x}_j$; conversely the same recursion expresses each power sum in them. The elementary symmetric polynomials are algebraically independent: in a lexicographic monomial order their products have distinct leading exponent lists, so a nonzero polynomial relation cannot cancel its leading monomial. They generate every symmetric polynomial as well: its lexicographically highest exponent list is nonincreasing, and a product of elementary symmetric polynomials has exactly that leading monomial; subtracting its scalar multiple and repeating strictly decreases the leading monomial within the finite degree. Their weighted degree-$n$ monomials, or equivalently the monomial symmetric functions indexed by partitions of $n$ with at most $m$ parts, thus give the required dimension-$\mathsf{p}(n)$ basis. The chain rule identifies $J_p$ with $\mathcal{J}=p{\sum }_i{y}_i^2{\mathrm{\partial }}_{y_i}^2+2{\sum }_{i<j}\frac{y_i^2{\mathrm{\partial }}_{y_i}-y_j^2{\mathrm{\partial }}_{y_j}}{y_i-y_j}-2(m-1){\sum }_i{y}_i{\mathrm{\partial }}_{y_i}\mathrm{.}$ To check the first-order terms, its action on $x_k$ is $k{\sum }_{a=1}^{k-1}{x}_a{x}_{k-a}+(p-1)k(k-1)x_k$. This follows from ${\sum }_{i<j}(y_i^{k+1}-y_j^{k+1})\mathrm{/}(y_i-y_j)=\frac{1}{2}({\sum }_{a=0}^k{x}_a{x}_{k-a}-(k+1)x_k)$ with $x_0=m$. The chain rule's second-order term is $p{\sum }_{a,b}ab{x}_{a+b}{\mathrm{\partial }}_{x_a}{\mathrm{\partial }}_{x_b}$, as displayed. The divided differences are polynomials on symmetric inputs since their numerators are antisymmetric in $y_i,y_j$.

In the monomial symmetric basis, $\mathcal{J}$ is triangular in partition dominance order, with diagonal entry ${\theta }_{\lambda }=p{\sum }_i{\lambda }_i({\lambda }_i-1)-2{\sum }_i(i-1){\lambda }_i\mathrm{.}$ For completeness, on two unequal exponents $a>b$ the pair operator $(y_i^2{\mathrm{\partial }}_{y_i}-y_j^2{\mathrm{\partial }}_{y_j})\mathrm{/}(y_i-y_j)$ sends $y_i^a{y}_j^b+y_i^b{y}_j^a$ to $a(y_i^a{y}_j^b+y_i^b{y}_j^a)$ plus terms with the two exponents strictly between $a$ and $b$. This is obtained by expanding the differences of powers in its numerator; the remaining terms transfer positive exponent from the larger to the smaller. Such transfers produce dominated partitions: the sum of any fixed number of the largest entries cannot increase, as follows by maximizing the sum over subsets before and after the transfer. For equal exponents the diagonal coefficient is $a$ and there are no transfer terms. Hence the pair terms have diagonal $2{\sum }_{i<j}\max ({\lambda }_i,{\lambda }_j)=2{\sum }_i(m-i){\lambda }_i$. Combining this with the first and last terms of $\mathcal{J}$ proves the formula. Further, ${\sum }_i(i-1){\lambda }_i\le n(n-1)\mathrm{/}2$, with equality only for $\lambda =(1^n)$: regarding the partition as ordered rows of boxes, the index $i-1$ counts preceding rows and is at most the number of boxes preceding a given box; summing gives the bound, strict if any row contains two boxes. Since $p>0$, it follows that ${\theta }_{\lambda }\ge -n(n-1)$, and equality holds only at $(1^n)$. Thus all diagonal entries of $-J_p+(2-p)n$ are strictly negative when $1\le n<N$, and at $n=N$ exactly one is zero, the column partition. By triangularity and conjugation, $H$ is invertible on those smaller degrees and its kernel has dimension at most one on degree $N$.

Induct on $n\le N$ for the span $D_n$. The degree-zero span is the whole one-dimensional space. Suppose all smaller degrees are full. By definition $D_n={\sum }_{k=1}^{n}L(-k)D_{n-k}$. Its annihilator under the nondegenerate pairing of step 1.1 is therefore precisely the simultaneous kernel of $L(1),\dots ,L(n)$ on the dual charge-$1\mathrm{/}2$ degree-$n$ space: the lower spaces are full by induction. Every vector in this simultaneous kernel is killed by $H$. Step 4.1 gives zero annihilator when $n<N$, hence $D_n=P_{n}w$. When $n=N$ it gives $\mathrm{codim}{D}_N\le 1$. On the other hand $Qw=0$, since $Y(e^{\alpha },z)w$ starts with $z^{p-2}b$, and [F3] implies that $Q$ kills every Virasoro descendant of $w$. Its explicit action on $P_{N}w$ is easy to calculate: the annihilation exponential replaces $x_k$ by $x_k-2p{z}^{-k}$, and the prefactor is $z^{p-2}$. The output charge $\beta +1$ has minimum degree $N$, so the zero-mode output here has oscillator degree zero. Only the constant output of that substitution contributes, giving $Q(fw)=f(-2p,-2p,\dots )b$ for weighted-homogeneous $f\in P_N$. This functional is nonzero, since its value on $x_N$ is $-2p$. Thus $D_N$ has codimension at least one and is contained in this rank-one kernel. The equal codimensions prove the asserted equality. Finally all negative Virasoro modes are modes of $\omega \in W(p)$, so $D_N\subset W(p)w$.